Optimal 8-vertex isoperimetric polyhedron? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T21:07:40Z http://mathoverflow.net/feeds/question/73899 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73899/optimal-8-vertex-isoperimetric-polyhedron Optimal 8-vertex isoperimetric polyhedron? Joseph O'Rourke 2011-08-28T14:48:14Z 2011-08-29T23:52:29Z <p>I know from Marcel Berger's <a href="http://www.springer.com/mathematics/geometry/book/978-3-540-70996-1" rel="nofollow"> <em>Geometry Revealed: A Jacob's Ladder to Modern Higher Geometry</em></a> (p.531) that it is not yet established which polyhedron in $\mathbb{R}^3$ on 8 vertices achieves the optimal isoperimetric ratio $A^3/V^2$, where $A$ is the surface area and $V$ the volume. Berger says, "We also know that the cube ... [is] not the best for $v=8$" (where $v$ is the number of vertices).</p> <p>Many other aspects of isoperimetry for polyhedra are unresolved, but this one especially interests me. It is not even clear to me that it is known that there <em>is</em> an optimal polyhedron for each $v$. I've been trying to imagine what would be a strong candidate for an optimal 8-vertex polyhedron. I've been unsuccessful in finding information on this, although it seems likely to have been explored computationally. Does anyone have a candidate, or know of one proposed/calculated? A pointer or reference would be greatly appreciated. Thanks!</p> <p><b>Addendum.</b> From the reference Igor provided (Nobuaki Mutoh, "The Polyhedra of Maximal Volume Inscribed in the Unit Sphere and of Minimal Volume Circumscribed about the Unit Sphere," 2009), here is a piece of Mutoh's Fig.1, which computationally verifies the earlier derivation of the max volume inscribed 8-vertex polyhedron by Berman and Haynes ("Volumes of polyhedra inscribed in the unit sphere in $\mathbb{R}^3$," <em>Math. Ann.</em>, <b>188</b>(1): 78-84, 1970), as mentioned in the comments: <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://csail.mit.edu/~orourke/MaxVol8.jpg" alt="MaxVol8"> <br /> This is surely a candidate for achieving the min of $A^3/V^2$! I thank Jean-Marc, Igor, and Anton for the rapid convergence to what I sought. </p> <p>...And then a bit later to Henry for showing that this candidate does not in fact achieve the best ratio! Here is Henry's polyhedron, if I have interpreted him correctly: <br /> <img src="http://cs.smith.edu/~orourke/MathOverflow/IsoCohn.jpg" alt="Cohn"></p> http://mathoverflow.net/questions/73899/optimal-8-vertex-isoperimetric-polyhedron/73941#73941 Answer by Henry Cohn for Optimal 8-vertex isoperimetric polyhedron? Henry Cohn 2011-08-29T05:14:12Z 2011-08-29T05:14:12Z <p>An $8$-vertex polyhedron can achieve an isoperimetric ratio of $A^3/V^2 = 159.3243297053\dots$, and based on some quick experiments I'm pretty confident this is optimal (although I wouldn't be shocked if it could be beaten).</p> <p>To construct it, let $V_\alpha$ denote the squashed tetrahedron with vertices $(\pm \sqrt{1-\alpha^2},0,\alpha)$ and $(0,\pm \sqrt{1-\alpha^2},-\alpha)$. Then the optimal $8$-vertex polyhedron seems to be the union of $V_\alpha$ and $-\beta V_\gamma$, with $\alpha = 0.2272117725\dots$, $\beta = 0.87345300464\dots$, and $\gamma = 0.83792301859\dots$. The optimal values of $\alpha$, $\beta$, and $\gamma$ are algebraic, but they're pretty complicated and I haven't computed their minimal polynomials.</p> <p>For comparison, the maximum volume polyhedron inscribed in a sphere has a worse isoperimetric ratio, namely $162.248792\dots$. For the cube, it's $216$.</p> <p>In general there's no reason to expect the optimal polyhedron to be inscribed in a sphere. The $5$-vertex case is a particularly nice example: it consists of an equilateral triangle on the equator of the unit sphere together with $1/\sqrt{2}$ times the north and south poles. This achieves an isoperimetric ratio of $243$, and I'd be very surprised if that's not optimal. Five vertices is few enough that a rigorous proof may be possible, but I can't think of a non-painful way to do it.</p>